# qr - Orthogonal Bases and the QR Algorithm by Peter J Olver...

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Unformatted text preview: Orthogonal Bases and the QR Algorithm by Peter J. Olver University of Minnesota 1. Orthogonal Bases. Throughout, we work in the Euclidean vector space V = R n , the space of column vectors with n real entries. As inner product, we will only use the dot product v · w = v T w and corresponding Euclidean norm bardbl v bardbl = √ v · v . Two vectors v , w ∈ V are called orthogonal if their inner product vanishes: v · w = 0. In the case of vectors in Euclidean space, orthogonality under the dot product means that they meet at a right angle. A particularly important configuration is when V admits a basis consisting of mutually orthogonal elements. Definition 1.1. A basis u 1 ,..., u n of V is called orthogonal if u i · u j = 0 for all i negationslash = j . The basis is called orthonormal if, in addition, each vector has unit length: bardbl u i bardbl = 1, for all i = 1 ,...,n . The simplest example of an orthonormal basis is the standard basis e 1 = 1 . . . , e 2 = 1 . . . , ... e n = . . . 1 . Orthogonality follows because e i · e j = 0, for i negationslash = j , while bardbl e i bardbl = 1 implies normality. Since a basis cannot contain the zero vector, there is an easy way to convert an orthogonal basis to an orthonormal basis. Namely, we replace each basis vector with a unit vector pointing in the same direction. Lemma 1.2. If v 1 ,..., v n is an orthogonal basis of a vector space V , then the normalized vectors u i = v i / bardbl v i bardbl , i = 1 ,...,n , form an orthonormal basis. Example 1.3. The vectors v 1 = 1 2 − 1 , v 2 = 1 2 , v 3 = 5 − 2 1 , are easily seen to form a basis of R 3 . Moreover, they are mutually perpendicular, v 1 · v 2 = v 1 · v 3 = v 2 · v 3 = 0, and so form an orthogonal basis with respect to the standard dot 6/5/10 1 c circlecopyrt 2010 Peter J. Olver product on R 3 . When we divide each orthogonal basis vector by its length, the result is the orthonormal basis u 1 = 1 √ 6 1 2 − 1 = 1 √ 6 2 √ 6 − 1 √ 6 , u 2 = 1 √ 5 1 2 = 1 √ 5 2 √ 5 , u 3 = 1 √ 30 5 − 2 1 = 5 √ 30 − 2 √ 30 1 √ 30 , satisfying u 1 · u 2 = u 1 · u 3 = u 2 · u 3 = 0 and bardbl u 1 bardbl = bardbl u 2 bardbl = bardbl u 3 bardbl = 1. The appearance of square roots in the elements of an orthonormal basis is fairly typical. A useful observation is that any orthogonal collection of nonzero vectors is automati- cally linearly independent....
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## This note was uploaded on 02/10/2012 for the course MATH 5485 taught by Professor Olver during the Fall '09 term at University of Central Florida.

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qr - Orthogonal Bases and the QR Algorithm by Peter J Olver...

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